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Physics MCQs for NEET — Practice Questions with Answers

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Kepler's second law regarding constancy of areal velocity of a planet is a consequence of law of conservation of

Energy

Linear momentum

Angular momentum

Mass

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Explanation

$A n g u l a r m o m e n t u m = 2 \times m a s s \times a r e a l s p e e d$

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A body of super dense material with mass twice the mass of the earth but size very small compared to size of the earth starts from rest from h<<R above the Earth's surface. It reaches earth in time t:

$t = \sqrt{\frac{h}{g}}$

$t = \sqrt{\frac{2 h}{g}}$

$t = \sqrt{\frac{2 h}{3 g}}$

$t = \sqrt{\frac{4 h}{3 g}}$

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Explanation

$F o r c e b e t w e e n t h e b o d y a n d e a r t h = \frac{2 G M^{2}}{R^{2}} A c c e l e r a t i o n o f t h e b o d y = \frac{F_{g}}{2 M} = \frac{2 G M^{2}}{2 M R^{2}} = \frac{G M}{R^{2}} = g A c c e l e r a t i o n o f t h e e a r t h = \frac{F_{g}}{M} = \frac{2 G M^{2}}{M R^{2}} = \frac{2 G M}{R^{2}} = 2 g A c c e l e r a t i o n o f t h e b o d y w . r . t e a r t h = g - (- 2 g) = 3 g T i m e = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 h}{3 g}}$

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A thin rod of length L is bent to form a semicircle. The mass of the rod is M. The gravitational potential at the centre of the circle is :

$- \frac{GM}{L}$

$- \frac{GM}{2 πL}$

$- \frac{πGM}{2 L}$

$- \frac{πGM}{L}$

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Explanation

$(4) As the rod is bent in the form of a semicircle, πR = L Radius of the circle, R = \frac{L}{π} V = - \frac{GM}{R} = - \frac{GM}{\frac{L}{π}} = - \frac{πGM}{L}$

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A point P lies on the axis of a ring of mass M and radius 'a' at a distance 'a' from its centre C. A small particle starts from P and reaches C under gravitational attraction. Its speed at C will be :

$\sqrt{\frac{2 GM}{a}}$

$\sqrt{\frac{2 GM}{a} (1 - \frac{1}{\sqrt{2}})}$

$\sqrt{\frac{2 GM}{a} (\sqrt{2} - 1)}$

zero

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Explanation

$(2) K_{p} + U_{p} = K_{c} + U_{c} 0 - \frac{GMm}{\sqrt{a^{2} + a^{2}}} = \frac{1}{2} {mv}^{2} - \frac{GMm}{a} v^{2} = 2 \times (- \frac{GMm}{\sqrt{2} a} + \frac{GMm}{a}) = \frac{2 GMm}{a} (- \frac{1}{\sqrt{2}} + 1) v = \sqrt{\frac{2 GMm}{a} (1 - \frac{1}{\sqrt{2}})}$

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Weightlessness experienced while orbiting the earth in space-ship, is the result of

Inertia

Acceleration

Zero gravity

Free fall towards earth

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Explanation

(d)

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The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km/sec

11.2

5.6

22.4

53.6

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Explanation

(c) $\frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{p}}{R_{e}}} = \sqrt{2 \times 2} = 2 \Rightarrow v_{p} = 2 \times v_{e} = 2 \times 11.2 = 22.4 k m / s$

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The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the earth, is

22 km/sec

11 km/sec

5.5 km/sec

15.5 km/sec

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Explanation

(a)

$v_{e} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 G}{R} \times \frac{4}{3} {πR}^{3} ρ} = R \sqrt{\frac{8}{3} πG ρ} S o i f p = c o n s \tan t, v_{e} α R . S i n c e t h e p l a n e t h a v i n g d o u b l e r a d i u s i n c o m p a r i s o n t o e a r t h, T h e r e f o r e t h e e s c a p e v e l o c i t y b e c o m e s t w i c e i . e . 22 k m / s .$

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What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

$7.4 \times 10^{- 4}$ rad/sac

6.4 $\times 10^{- 4}$ rad/sac

$7.8 \times 10^{- 4}$ rad/sac

8.7 $\times 10^{- 4}$ rad/sac

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Explanation

$W e i g h t o f t h e b o d y a t e q u a t o r = \frac{3}{5} o f i n i t i a l w e i g h t s o g' = \frac{3}{5} g (b e c a u s e m a s s r e m a i n s c o n s \tan t) g' = g - ω^{2} R \cos^{2} θ \Rightarrow \frac{3}{5} g = g - ω^{2} R \cos^{2} (0^{0}) \Rightarrow ω^{2} = \frac{2 g}{5 R} \Rightarrow w = \sqrt{\frac{2 g}{5 R}} = \sqrt{\frac{2 \times 10}{5 \times 6400 \times 10^{3}}} = 7.8 \times 10^{- 4} \frac{r a d}{s a c}$

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If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is

$\sqrt{2 g r}$

C.
g/r
D.

r/g

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Explanation

(b)

$K E + P E = 0 \frac{1}{2} m v^{2} - \frac{G M m}{r} = 0 \Rightarrow v = \sqrt{\frac{2 G M}{r}} O r, v = \sqrt{2 g r} (a s g = \frac{G M}{r^{2}})$

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The escape velocity of a projectile from the earth is approximately

11.2 m/sec

112 km/sec

11.2 km/sec

11200 km/sec

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Explanation

(c)

The escape velocity of a projectile from the earth is approximately 11.2 Km/s.

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.